Question
A chemist prepares a solution of mercuryII) iodide (Hgi, by measuring out 0.011 g of mercuryIl) iodide into a 250. mi volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's mercury(II) iodide solution. Be sure your answer has the correct number of significant digits.
Calculate the concentration in mol/L of the chemist's mercury(II) iodide solution. Be sure your answer has the correct number of significant digits.
Answers
GPT 3.5
To calculate the concentration of the mercury(II) iodide solution, we first need to calculate the number of moles of HgI2 present in the solution.
Molar mass of HgI2:
Hg = 200.59 g/mol
I = 126.90 g/mol
HgI2 = 200.59 g/mol + 2(126.90 g/mol) = 454.39 g/mol
Number of moles of HgI2 = 0.011 g / 454.39 g/mol = 2.42 x 10^-5 mol
Now, we need to calculate the volume of the solution in liters:
250. ml = 250. x 10^-3 L = 0.250 L
Concentration of HgI2 = number of moles / volume of solution
Concentration = 2.42 x 10^-5 mol / 0.250 L = 9.68 x 10^-5 mol/L
Therefore, the concentration of the chemist's mercury(II) iodide solution is 9.68 x 10^-5 mol/L.
Molar mass of HgI2:
Hg = 200.59 g/mol
I = 126.90 g/mol
HgI2 = 200.59 g/mol + 2(126.90 g/mol) = 454.39 g/mol
Number of moles of HgI2 = 0.011 g / 454.39 g/mol = 2.42 x 10^-5 mol
Now, we need to calculate the volume of the solution in liters:
250. ml = 250. x 10^-3 L = 0.250 L
Concentration of HgI2 = number of moles / volume of solution
Concentration = 2.42 x 10^-5 mol / 0.250 L = 9.68 x 10^-5 mol/L
Therefore, the concentration of the chemist's mercury(II) iodide solution is 9.68 x 10^-5 mol/L.